Bi-wired Booshelf Speakers

[jdh]

Hello Folks,

I just bought a used pair of bookshelf speaekers.

Each cabinet has two pairs of speaker wires, one for what the decal calls the "speaker," rated at 4 Ohms, the other for the "subwoofer," ratend at 6 Ohms. (The term "subwoofer" seems to refer to an eight inch unit. Now, now...)

I'm installing these speakers in a very small room, powering them with a two channel receiver.

Is it as simple as connecting the two negaitve wires of each cabinet to each other, doing the same for the possitives, and treating the cabinet as if it's wired "normaly"?

Is there a way to calculate or guess what the new impedence would be?

Thanks in advance.

JDH

 

[j_garcia]

I believe you must wire them in parallel (+ to +, - to -) for any internal crossover to function, so that will be your only choice. This is not biwiring, this would be "normal wiring" as you said, and this is what I would most likely recommend.

Here is the formula for calculating impedance for parallel wiring:

Zt = (Za x Zb) / (Za + Zb)

SO:
Zt = (6 x 4) / (6 + 4)
Zt = 24 / 10
Zt = 2.4 ohms

That's very low.

 

[Swerd]

I believe you must wire them in parallel (+ to +, - to -) for any internal crossover to function, so that will be your only choice. This is not biwiring, this would be "normal wiring" as you said, and this is what I would most likely recommend.

j_garcia
Either I don't understand your wording, or this is incorrect.

jdh
Do the terminals on the back of your speakers (are these speakers 2-ways?) look something like the picture? Two sets of terminals mean that the crossover circuits for the woofer and tweeter are electrically separated from each other instead of sharing common + and - inputs. You have two choices of connecting them.

On the speaker, you can join one + to the other + with a short jumper wire, and the same with the two - terminals. In the picture they are already joined with a brass jumper bar. Your speakers may have once had those, but the jumper wires will work the same. Then run zipcord from your amp to the speakers in the usual way with one wire going from the + terminal on the amp to either of the + plus terminals on your speaker, and the other wire going from - on the amp to either of the - terminals on the speaker.

The other way is the so-called bi-wire method. You need two lengths of wire for each speaker, but do NOT use any jumpers. Join two wires together at the + amp terminal, and two wires at the - amp terminal for each speaker. Run a pair to each of the + and - terminals. Electrically this is the same as the first method except you double the amount of copper wire. There is considerable debate whether bi-wiring offers any audible benefit. Let's sidestep that debate because there is no clear answer.

The idea behind using separate terminals for the woofer and tweeter is that it also allows bi-amping - the use of two separate amps (or amp channels) one for the woofer and one for the tweeter. Obviously it costs a lot more, but there is little debate about the benefits of doing this.

 

[mtrycrafts]

Hello Folks,

I just bought a used pair of bookshelf speaekers.

Each cabinet has two pairs of speaker wires, one for what the decal calls the "speaker," rated at 4 Ohms, the other for the "subwoofer," ratend at 6 Ohms. (The term "subwoofer" seems to refer to an eight inch unit. Now, now...)

I'm installing these speakers in a very small room, powering them with a two channel receiver.

Is it as simple as connecting the two negaitve wires of each cabinet to each other, doing the same for the possitives, and treating the cabinet as if it's wired "normaly"?

Is there a way to calculate or guess what the new impedence would be?

Thanks in advance.

JDH

Do you have a brand name for this speaker? Does it actually have speaker wires coming from it or terminals as in swerds picture? Do you see a jumper between the same colored terminals?
The speakers impedance, is what the maker determines it to be. The impedance calculations above is incorrect if these are two drivers with crossover networks as the low driver will have very high impedance to the highs and the other way for the other driver.

 

[zumbo]

Multimeter. Some have argued this is not accurate, but it is a method I have always used. Test the top. Then the bottom. Then strapped.

http://i19.photobucket.com/albums/b152/tpc3416/9-4-05041.jpg

http://i19.photobucket.com/albums/b152/tpc3416/9-4-05042.jpg

http://i19.photobucket.com/albums/b152/tpc3416/9-4-05040.jpg

 

[mtrycrafts]

Multimeter. Some have argued this is not accurate, but it is a method I have always used. Test the top. Then the bottom. Then strapped.

That gives DC resistance not what the amp sees, frequency dependent impedance. At the low end, the high frequency has very high impedance and that is what matters, not the DC resistance of the high driver.
That speaker in the link can have the lowes impedance of 7.1 ohms, the lowes driver DC resistance, not their parallel resistance.

 

[jdh]

Hello Folks,

Thank you so much for the great replies.

I'm taking the speakers apart in the morning to check how everything is connected.

However...

There are no connections of any sort anywhere -- just the four wires coming out of each cabinet.

Regards,

JDH

 

[mtrycrafts]

Hello Folks,

Thank you so much for the great replies.

I'm taking the speakers apart in the morning to check how everything is connected.

However...

There are no connections of any sort anywhere -- just the four wires coming out of each cabinet.

Regards,

JDH

Just as I gathered from your post. So, it seems that they are rather cheap speakers? Brand and model?
Yes, you looking inside may shed some light how this is wired, or, who knows what is in there

 

[jdh]

Hello Everyone,

Just as I was about to open up these things, the seller agreed to take them back.

Someone said these may be cheap speakers.

Ooh, yeah.

JVC Megatubes!

JDH